Question

54 simplify the radical $-8\sqrt{63}$

Explanation:

Step1: Factor the radicand

Factor 63 into prime factors: \(63 = 9\times7\), where 9 is a perfect square. So, \(\sqrt{63}=\sqrt{9\times7}\).

Step2: Apply the square - root property

Using the property \(\sqrt{ab}=\sqrt{a}\times\sqrt{b}\) (\(a\geq0,b\geq0\)), we have \(\sqrt{9\times7}=\sqrt{9}\times\sqrt{7}\). Since \(\sqrt{9} = 3\), then \(\sqrt{9}\times\sqrt{7}=3\sqrt{7}\).

Step3: Multiply by the coefficient

We have \(- 8\sqrt{63}=-8\times3\sqrt{7}\).

Step4: Calculate the product of coefficients

\(-8\times3=-24\), so \(-8\sqrt{63}=-24\sqrt{7}\).

Answer:

\(-24\sqrt{7}\)